Problem: Recently Duff has been a soldier in the army. Malek is her commander.

Their country, Andarz Gu has n cities (numbered from 1 to n) and n - 1 bidirectional roads. Each road connects two different cities. There exist a unique path between any two cities.

There are also m people living in Andarz Gu (numbered from 1 to m). Each person has and ID number. ID number of i - th person is i and he/she lives in city number ci. Note that there may be more than one person in a city, also there may be no people living in the city.

<image>

Malek loves to order. That's why he asks Duff to answer to q queries. In each query, he gives her numbers v, u and a.

To answer a query:

Assume there are x people living in the cities lying on the path from city v to city u. Assume these people's IDs are p1, p2, ..., px in increasing order. 

If k = min(x, a), then Duff should tell Malek numbers k, p1, p2, ..., pk in this order. In the other words, Malek wants to know a minimums on that path (or less, if there are less than a people).

Duff is very busy at the moment, so she asked you to help her and answer the queries.

Input

The first line of input contains three integers, n, m and q (1 ≤ n, m, q ≤ 105).

The next n - 1 lines contain the roads. Each line contains two integers v and u, endpoints of a road (1 ≤ v, u ≤ n, v ≠ u).

Next line contains m integers c1, c2, ..., cm separated by spaces (1 ≤ ci ≤ n for each 1 ≤ i ≤ m).

Next q lines contain the queries. Each of them contains three integers, v, u and a (1 ≤ v, u ≤ n and 1 ≤ a ≤ 10).

Output

For each query, print numbers k, p1, p2, ..., pk separated by spaces in one line.

Examples

Input

5 4 5
1 3
1 2
1 4
4 5
2 1 4 3
4 5 6
1 5 2
5 5 10
2 3 3
5 3 1


Output

1 3
2 2 3
0
3 1 2 4
1 2

Note

Graph of Andarz Gu in the sample case is as follows (ID of people in each city are written next to them):

<image>

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
void printvec(vector<int> v, int a) {
  cout << min(a, (int)v.size()) << ' ';
  for (int i = 0; i < a && i < v.size(); i++) {
    cout << v[i] << ' ';
  }
  cout << endl;
}
const int N = 1e5 + 564;
const int LOG = 20;
int c[N];
int up[N][LOG + 5];
vector<int> vec[N][LOG + 6], ord[N];
vector<int> g[N];
int level[N];
vector<int> Merge(vector<int> a, vector<int> b) {
  for (int i = 0; i < b.size(); i++) {
    a.push_back(b[i]);
  }
  sort(a.begin(), a.end());
  vector<int> res;
  for (int i = 0; i < min((int)a.size(), 10); i++) {
    res.push_back(a[i]);
  }
  return res;
}
void dfs(int v = 1, int p = 0) {
  vec[v][0] = ord[p];
  up[v][0] = p;
  for (int i = 1; i <= LOG; i++) {
    up[v][i] = up[up[v][i - 1]][i - 1];
    vec[v][i] = Merge(vec[v][i - 1], vec[up[v][i - 1]][i - 1]);
  }
  for (auto to : g[v]) {
    if (to != p) {
      level[to] = level[v] + 1;
      dfs(to, v);
    }
  }
}
vector<int> lca(int a, int b) {
  if (level[a] < level[b]) {
    swap(a, b);
  }
  vector<int> resa = ord[a], resb = ord[b];
  for (int i = LOG; i >= 0; i--) {
    if (up[a][i] != 0 && level[up[a][i]] >= level[b]) {
      resa = Merge(resa, vec[a][i]);
      a = up[a][i];
    }
  }
  if (a == b) {
    return resa;
  }
  for (int i = LOG; i >= 0; i--) {
    if (up[a][i] != 0 && up[a][i] != up[b][i]) {
      resa = Merge(resa, vec[a][i]);
      resb = Merge(resb, vec[b][i]);
      a = up[a][i];
      b = up[b][i];
    }
  }
  resa = Merge(resa, ord[up[a][0]]);
  resa = Merge(resa, resb);
  return resa;
}
int main() {
  int n, m, q;
  cin >> n >> m >> q;
  for (int i = 1; i < n; i++) {
    int a, b;
    cin >> a >> b;
    g[a].push_back(b);
    g[b].push_back(a);
  }
  for (int i = 1; i <= m; i++) {
    cin >> c[i];
    if (ord[c[i]].size() < 10) {
      ord[c[i]].push_back(i);
    }
  }
  dfs();
  while (q--) {
    int v, u, a;
    cin >> v >> u >> a;
    printvec(lca(v, u), a);
  }
  return 0;
}